Conditional probablility P(A/B) = P(A and B) / P(B). Here, A is sum of two dice being greater than or equal to 9 and B is at least one of the dice showing 6. Number of ways two dice faces can sum up to 9 = (3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6) = 10 ways. Number of ways that at least one of the dice must show 6 = (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6), (6, 5), (6, 4), (6, 3), (6, 2), (6, 1) = 11 ways. Number of ways of rolling a number greater than or equal to 9 and at least one of the dice showing 6 = (3, 6), (4, 6), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6) = 7 ways. Probability of rolling a number greater than or equal to 9 given that at least one of the dice must show a 6 = 7 / 11
Answer:
<h2>
46</h2>
Step-by-step explanation:
x - age of a son three years ago
3x - age of a father three years ago
x + 3 - age of the son now
3x + 3 - age of the father now
x + 3 + 5 - age of the son in five years time
3x + 3 + 5 - age of the father in five years time
in five years time, the father will be twice as old as his son, so:
3x + 3 + 5 = 2(x + 3 + 5)
3x + 8 = 2x + 16
3x - 2x = 16 - 8
x = 8
x + 3 + 4 - age of the son in four years time
8 + 3 + 4 = 15
3x + 3 + 4 - age of the father in four years time
3×8 + 3 + 4 = 31
the sum of their ages in four years time:
15 + 31 = 46